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c - C 中删除列表节点的这一段是什么意思

转载 作者:行者123 更新时间:2023-11-30 16:09:38 25 4
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下面是删除链接简单列表的最后一个节点的完整代码,如果运行良好,函数为 void deleteLastNode ()。

但这部分到底是什么意思呢?在我看来,这是多余的,因为当 while 循环结束时,我们将 toDelete 节点放在最后一个位置,将 secondaryLastNode 放在倒数第二个位置。

if(toDelete == head)
{
head = NULL;
}

deleteLastNode () 函数是这样的:

void deleteLastNode()
{
struct node *toDelete, *secondLastNode;

if(head == NULL)
{
printf("List is already empty.");
}
else
{
toDelete = head;
secondLastNode = head;

/* Traverse to the last node of the list */
while(toDelete->next != NULL)
{
secondLastNode = toDelete;
toDelete = toDelete->next;
}

if(toDelete == head)
{
head = NULL;
}
else
{
/* Disconnect link of second last node with last node */
secondLastNode->next = NULL;
}

/* Delete the last node */
free(toDelete);

printf("SUCCESSFULLY DELETED LAST NODE OF LIST\n");
}
}

但我相信这已经足够了:

/*  */
void deleteLastNode()
{
struct node *toDelete, *secondLastNode;

if(head == NULL)
{
printf("List is already empty.");
}
else
{
toDelete = head;
secondLastNode = head;

/* Traverse to the last node of the list */
while(toDelete->next != NULL)
{
secondLastNode = toDelete;
toDelete = toDelete->next;
}





/* Disconnect link of second last node with last node */
secondLastNode->next = NULL;


/* Delete the last node */
free(toDelete);

printf("SUCCESSFULLY DELETED LAST NODE OF LIST\n");
}
}

完整的代码是这样的:函数是void deleteLastNode()::

/**
* C program to delete last node of Singly Linked List
*/

#include <stdio.h>
#include <stdlib.h>


/* Structure of a node */
struct node {
int data; // Data
struct node *next; // Address
}*head;

void createList(int n);
void deleteLastNode();
void displayList();


int main()
{
int n, choice;

/*
* Create a singly linked list of n nodes
*/
printf("Enter the total number of nodes: ");
scanf("%d", &n);
createList(n);

printf("\nData in the list \n");
displayList();

printf("\nPress 1 to delete last node: ");
scanf("%d", &choice);

/* Delete last node from list */
if(choice == 1)
deleteLastNode();

printf("\nData in the list \n");
displayList();

return 0;
}


/*
* Create a list of n nodes
*/
void createList(int n)
{
struct node *newNode, *temp;
int data, i;

head = (struct node *)malloc(sizeof(struct node));

/*
* If unable to allocate memory for head node
*/
if(head == NULL)
{
printf("Unable to allocate memory.");
}
else
{
/*
* Input data of node from the user
*/
printf("Enter the data of node 1: ");
scanf("%d", &data);

head->data = data; // Link the data field with data
head->next = NULL; // Link the address field to NULL

temp = head;

/*
* Create n nodes and adds to linked list
*/
for(i=2; i<=n; i++)
{
newNode = (struct node *)malloc(sizeof(struct node));

/* If memory is not allocated for newNode */
if(newNode == NULL)
{
printf("Unable to allocate memory.");
break;
}
else
{
printf("Enter the data of node %d: ", i);
scanf("%d", &data);

newNode->data = data; // Link the data field of newNode with data
newNode->next = NULL; // Link the address field of newNode with NULL

temp->next = newNode; // Link previous node i.e. temp to the newNode
temp = temp->next;
}
}

printf("SINGLY LINKED LIST CREATED SUCCESSFULLY\n");
}
}


/*
* Delete last node of the linked list
*/

void deleteLastNode()
{
struct node *toDelete, *secondLastNode;

if(head == NULL)
{
printf("List is already empty.");
}
else
{
toDelete = head;
secondLastNode = head;

/* Traverse to the last node of the list */
while(toDelete->next != NULL)
{
secondLastNode = toDelete;
toDelete = toDelete->next;
}

if(toDelete == head)
{
head = NULL;
}
else
{
/* Disconnect link of second last node with last node */
secondLastNode->next = NULL;
}

/* Delete the last node */
free(toDelete);

printf("SUCCESSFULLY DELETED LAST NODE OF LIST\n");
}
}


/*
* Display entire list
*/
void displayList()
{
struct node *temp;

/*
* If the list is empty i.e. head = NULL
*/
if(head == NULL)
{
printf("List is empty.");
}
else
{
temp = head;
while(temp != NULL)
{
printf("Data = %d\n", temp->data); // Print the data of current node
temp = temp->next; // Move to next node
}
}
}

谢谢

最佳答案

在第二个函数实现中,如果最后一个节点是头节点,则变量 head 本身不会更改。因此,当列表仅包含一个作为头节点的节点时,该函数可能会出现未定义的行为。

请注意,无论如何函数实现都是不好的。例如,该函数处理全局变量 head。所以在一个程序中你不能定义两个列表。

该函数可以写得更简单。

void deleteLastNode( struct node **head )
{
if ( *head != NULL )
{
while ( ( *head )->next != NULL ) head = &( *head )->next;

free( *head );

*head = NULL;
}
}

并称其为

deleteLastNode( &head );

关于c - C 中删除列表节点的这一段是什么意思,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/59095992/

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